On Tue, Aug 28, 2012 at 5:54 PM, Chris Smith <[email protected]> wrote: >>>> from sympy.combinatorics import * >>>> Cycle()*(1,2)*(2,3) > [(1, 3, 2)]
I call this L-R multiplication, because you "plug" 1 in from the left and see what cycle it belongs to by scanning L to R, then plug in the next smallest integer outside that cycle and see what cycle it belongs to, etc This agrees with Sage and Gap: sage: G = SymmetricGroup(5) sage: g1 = G([(1,2)]) sage: g2 = G([(2,3)]) sage: g1*g2 (1,3,2) >>>> _.as_list() > [0, 3, 1, 2] >>>> Permutation([[2,3]],size=4).array_form > [0, 1, 3, 2] >>>> Permutation([[1,2]],size=4).array_form > [0, 2, 1, 3] >>> Permutation([[1,2]],size=4)*Permutation([[2,3]],size=4) > Permutation([0, 2, 3, 1]) Which is (2,1,3). > > So using the (1,2) permutation to select from the (2,3) one gives the > final answer of [0, 3, 1, 2] (applied the (1,2) last, hence R to L) > > -- > You received this message because you are subscribed to the Google Groups > "sympy" group. > To post to this group, send email to [email protected]. > To unsubscribe from this group, send email to > [email protected]. > For more options, visit this group at > http://groups.google.com/group/sympy?hl=en. > -- You received this message because you are subscribed to the Google Groups "sympy" group. To post to this group, send email to [email protected]. To unsubscribe from this group, send email to [email protected]. For more options, visit this group at http://groups.google.com/group/sympy?hl=en.
